Maximum Range of a Projectile

To determine the maximum range of a projectile, we need to understand the factors that affect its motion. When a particle is projected from a point on the ground with an initial velocity u, it follows a curved path known as a projectile motion. The range of the projectile is the horizontal distance covered by the particle before it hits the ground again.

1. Factors Affecting Projectile Motion:
- Initial velocity (u): The magnitude and direction of the initial velocity of the particle will determine the shape of the projectile's trajectory.
- Angle of projection (θ): The angle at which the particle is launched affects both the vertical and horizontal components of its motion.
- Acceleration due to gravity (g): Gravity acts in the vertical direction and influences the vertical component of the projectile's motion.

2. Maximum Range:
- The maximum range of a projectile occurs when the angle of projection is 45 degrees. At this angle, the horizontal and vertical components of the initial velocity are equal.
- The horizontal distance covered by the particle can be calculated using the formula: R = (u^2 * sin(2θ))/g.
- To maximize the range, we need to find the angle that maximizes the value of sin(2θ).
- The maximum value of sin(2θ) is 1, which occurs when 2θ = 90 degrees or θ = 45 degrees.

3. Average Velocity:
- The average velocity of the projectile during its flight can be calculated by dividing the total displacement by the total time taken.
- In the case of maximum range, the displacement in the horizontal direction is equal to the range (R).
- The total time taken can be calculated using the formula: t = 2u*sin(θ)/g.
- Therefore, the average velocity (Vavg) is given by Vavg = R/t = R/(2u*sin(θ)/g) = (R * g)/(2u*sin(θ)).

In conclusion, to achieve the maximum range for a projectile, it should be launched at an angle of 45 degrees. At this angle, the magnitude of the average velocity during its flight can be calculated using the formula (R * g)/(2u*sin(θ)).